Question

Calculate the molarity of each of the following ions. 4th attempt Part 1 (1 point) 0.933 g Nat in 130.0 mL of a NaCl solution. Part 2 (1 point) 0.760 g of Cl in a 130.0 mL of a NaCl solution.

Answer 1

1)
Molar mass of Na = 22.99 g/mol


mass(Na)= 0.933 g

use:
number of mol of Na,
n = mass of Na/molar mass of Na
=(0.933 g)/(22.99 g/mol)
= 4.058*10^-2 mol
volume , V = 1.3*10^2 mL
= 0.13 L


use:
Molarity,
M = number of mol / volume in L
= 4.058*10^-2/0.13
= 0.3122 M
Answer: 0.312 M

2)
Molar mass of Cl = 35.45 g/mol


mass(Cl)= 0.760 g

use:
number of mol of Cl,
n = mass of Cl/molar mass of Cl
=(0.76 g)/(35.45 g/mol)
= 2.144*10^-2 mol
volume , V = 1.3*10^2 mL
= 0.13 L


use:
Molarity,
M = number of mol / volume in L
= 2.144*10^-2/0.13
= 0.1649 M
Answer: 0.165 M

3)
Molar mass of SO4,
MM = 1*MM(S) + 4*MM(O)
= 1*32.07 + 4*16.0
= 96.07 g/mol


mass(SO4)= 0.400 g

use:
number of mol of SO4,
n = mass of SO4/molar mass of SO4
=(0.4 g)/(96.07 g/mol)
= 4.164*10^-3 mol
volume , V = 1.3*10^2 mL
= 0.13 L


use:
Molarity,
M = number of mol / volume in L
= 4.164*10^-3/0.13
= 3.203*10^-2 M
Answer: 3.20*10^-2 M

4)
Molar mass of Ca = 40.08 g/mol


mass(Ca)= 0.230 g

use:
number of mol of Ca,
n = mass of Ca/molar mass of Ca
=(0.23 g)/(40.08 g/mol)
= 5.739*10^-3 mol
volume , V = 1.3*10^2 mL
= 0.13 L


use:
Molarity,
M = number of mol / volume in L
= 5.739*10^-3/0.13
= 4.414*10^-2 M
Answer: 4.41*10^-2 M